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Inessa [10]
2 years ago
15

After carrying out an ANOVA procedure where the decision is made to reject the null hypothesis, we can test for differences betw

een treatment means by ______.
Mathematics
1 answer:
aleksandrvk [35]2 years ago
4 0

Answer:

Doing an additional ANOVA

Explanation:

Analysis of variance(ANOVA), developed by Ronald Fisher is a method used to statistically measure the difference between means of different variables. While the t test measures difference between two population means, analysis of variance measures the difference for more than two population means. There is the one way ANOVA and two way ANOVA that test the difference between means.

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The length of a rectangle is 6 ft longer than its width. if the perimeter of the rectangle is 52 ft , find its area.
sveticcg [70]
Let its width be w and length= w+6
2(w+6+w)=52
4w+12=52
4w=40
w=10
Area=10*(10+6)=160
7 0
3 years ago
Find the center and radius of the circle (x+3)^2+(y-1)^2=81
sasho [114]

The equation of a circle is written as ( x-h)^2 + (y-k)^2 = r^2

h and k is the center point of the circle and r is the radius.

In the given equation (x+3)^2 + (y-1)^2 = 81

h = -3

k = 1

r^2 = 81

Take the square root of both sides:

r = 9

The center is (-3,1) and the radius is 9

5 0
3 years ago
Retail value is also known as the
Setler79 [48]

Answer:

Retail value is also known as the sticker value.

6 0
2 years ago
Read 2 more answers
Help please.
FrozenT [24]

Answer:

B think

Step-by-step explanation:

4 0
2 years ago
Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview
svetlana [45]

Answer:  The required solution is

y=(-2+t)e^{-5t}.

Step-by-step explanation:   We are given to solve the following differential equation :

y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.

So, the general solution of the given equation is

y(t)=(A+Bt)e^{-5t}.

Differentiating with respect to t, we get

y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.

According to the given conditions, we have

y(0)=-2\\\\\Rightarrow A=-2

and

y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.

Thus, the required solution is

y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.

6 0
3 years ago
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